Hive JDBC Kerberos连接错误 [英] Hive JDBC Kerberos Connectivity error

问题描述

我正在尝试通过JDBC连接到Hive-Server2.服务器已启用kerberos.

I am trying to connect to Hive-Server2 via JDBC.Server is kerberos enabled.

我正在使用以下代码，正在运行存在keytab文件的边缘节点的代码

I am using the following code,I am running the code of an edge node where keytab file is present

我想念什么?

import org.apache.hadoop.security.UserGroupInformation;

public class HiveJdbcClient {
private static String driverName = "org.apache.hive.jdbc.HiveDriver";

public static void main(String[] args)
        throws SQLException, IOException, ClassNotFoundException {

    org.apache.hadoop.conf.Configuration conf = new org.apache.hadoop.conf.Configuration();
    conf.set("hadoop.security.authentication", "Kerberos");
    UserGroupInformation.setConfiguration(conf);

    int ch = 1;
    String USER = (ch == 0) ? "hive/_HOST@HADOOP_DEV9.xyz.COM"
            : "abcd@HADOOP_DEV9.xyz.COM";
    String PASSWORD = (ch == 0) ? "/etc/security/keytabs/hive.service.keytab"
            : "/u/users/abcd/.abcd.keytab";
    System.out.println("User = " + USER + "Keytab =" + PASSWORD);

    File f = new File(PASSWORD);
    if (f.exists()) {
        System.out.println("File exists");
    } else {
        System.out.println("File not exists");
    }
    UserGroupInformation.loginUserFromKeytab(USER, PASSWORD);

    String ConnectionURL = "jdbc:hive2://tstr400382.xyz.com:2181,tstr400383.xyz.com:2181,tstr400384.xyz.com:2181,tstr400386.xyz.com:2181,tstr400387.xyz.com:2181/;"
            + "serviceDiscoveryMode=zooKeeper;zooKeeperNamespace=hiveserver2"
            + ";principal=hive/_HOST@HADOOP_DEV9.xyz.COM;transportMode=http;httpPath=cliservice";

    System.out.println("below connection url");

    Class.forName(driverName);
    Connection con = DriverManager.getConnection(ConnectionURL);
    Statement stmt = con.createStatement();
    String sql = "show databases;";
    System.out.println("Running: " + sql);
    ResultSet res = stmt.executeQuery(sql);
    if (res.next()) {
        System.out.println(res.getString(1));
    }

}
}

我遇到以下错误

Exception in thread "main" java.io.IOException: Login failure for abcd@HADOOP_DEV9.xyz.COM from keytab /u/users/abcd/.abcd.keytab: javax.security.auth.login.LoginException: Unable to obtain password from user

at org.apache.hadoop.security.UserGroupInformation.loginUserFromKeytab(UserGroupInformation.java:1098)
at hadoop.test.HiveJdbcClient.main(HiveJdbcClient.java:42)
 Caused by: javax.security.auth.login.LoginException: Unable to obtain password from user

at com.sun.security.auth.module.Krb5LoginModule.promptForPass(Krb5LoginModule.java:897)
at com.sun.security.auth.module.Krb5LoginModule.attemptAuthentication(Krb5LoginModule.java:760)
at com.sun.security.auth.module.Krb5LoginModule.login(Krb5LoginModule.java:617)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:483)
at javax.security.auth.login.LoginContext.invoke(LoginContext.java:755)
at javax.security.auth.login.LoginContext.access$000(LoginContext.java:195)
at javax.security.auth.login.LoginContext$4.run(LoginContext.java:682)
at javax.security.auth.login.LoginContext$4.run(LoginContext.java:680)
at java.security.AccessController.doPrivileged(Native Method)
at javax.security.auth.login.LoginContext.invokePriv(LoginContext.java:680)
at javax.security.auth.login.LoginContext.login(LoginContext.java:587)
at org.apache.hadoop.security.UserGroupInformation.loginUserFromKeytab(UserGroupInformation.java:1089)

我能够做到

kinit -kt /u/users/abcd/.abcd.keytab abcd@HADOOP_DEV9.xyz.COM

密钥表上的Klist

Klist on keytab

Keytab name: FILE:.abcd.keytab
   KVNO Principal
---- --------------------------------------------------------------------------
   1 abcd@HADOOP_DEV9.xyz.COM
   1 abcd@HADOOP_DEV9.xyz.COM

我长期坚持使用它，请帮忙.

I am stuck on it from a long time,please help.

我使用了

I used this for learning and I followed it

推荐答案

用于服务主体"hive"和 user主体的Kerberos领域名称可能有所不同"abcd"，如下所述:

There could be a difference in the Kerberos realm names used for the service principal "hive" and user principal "abcd" as mentioned below:

hive/_HOST@<KDC_REALM_1>

abcd@<KDC_REALM_2>

请与您的集群管理员确认，并进行必要的更改.

Please verify with your cluster administrator and make the necessary changes if there is a difference.

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